Difference between revisions of "2021 AIME I Problems/Problem 7"
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<math>mx \equiv nx \equiv \frac{\pi}{2} (</math>mod <math>2\pi).</math> | <math>mx \equiv nx \equiv \frac{\pi}{2} (</math>mod <math>2\pi).</math> | ||
− | This means that <math>mx = \frac{\pi}{2} + 2\pi\alpha</math> and <math>nx = \frac{\pi}{2} + 2\pi\beta</math> | + | This means that <math>mx = \frac{\pi}{2} + 2\pi\alpha</math> and <math>nx = \frac{\pi}{2} + 2\pi\beta</math> for any integers <math>\alpha</math> and <math>\beta</math>. |
I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN THANK YOU! | I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN THANK YOU! |
Revision as of 03:05, 13 March 2021
Contents
Problem
Find the number of pairs of positive integers with
such that there exists a real number
satisfying
Solution 1
The maximum value of is
, which is achieved at
for some integer
. This is left as an exercise to the reader.
This implies that , and that
and
, for integers
.
Taking their ratio, we have
It remains to find all
that satisfy this equation.
If , then
. This corresponds to choosing two elements from the set
. There are
ways to do so.
If , by multiplying
and
by the same constant
, we have that
. Then either
, or
. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set
. There are
ways here.
Finally, if , note that
must be an integer. This means that
belong to the set
, or
. Taking casework on
, we get the sets
. Some sets have been omitted; this is because they were counted in the other cases already. This sums to
.
In total, there are pairs of
.
This solution was brought to you by ~Leonard_my_dude~
Solution 2
In order for ,
.
This happens when
mod
This means that and
for any integers
and
.
I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN THANK YOU!
-KingRavi
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.